Any implications of this diagonal element inequality?

[winterfors]

Given two definite positive definite matrices A and B of identical size with the following relationship of their diagonal elements:

A_{ii} \geq B_{ii} (no summation)

which also holds after any unitary change of coordinates

\textbf{A}'=\textbf{U}^T\textbf{A}\textbf{U}

where \textbf{U} complete set of orthonormal vectors.


Question: What does this imply in terms of inequalities on the eigenvalues of the matrices?

The sum of eigenvalues of A is of course equal of greater than the sum of eigenvalues B:s eigenvalues, but this is true even without allowing for change of coordinates. I'm sure you must be able to deduce something stronger when the inequality holds under any unitary coordinate change...

 

[acarchau]

My linear algebra is very rusty so be careful of the argument below.

A-B is p.s.d .

I am assuming the matrices are real.

Now from the given condition we have for any orthogonal matrix U, the diagonal elements of U^T (A-B) U are non negative.

In particular choose U to be the orthogonal matrix corresponding to the spectral decomposition of A-B and we get all its eigenvalues as non negative.

 

[acarchau]

I think this also implies that the eigenvalues of A dominate the eigenvalues of B in the following way: if \lambda_i(A) is the i^{th} largest eigenvalue of A and \lambda_i(B) is similarly defined. Then \sum_{i=1}^{k} \lambda_i(A) \geq \sum_{i=1}^{k} \lambda_i(B) for k=1,\dots,n.

This follows from the fact that for two hermitian matrices A and B of the same order we have (4.3.27 Horn and Johnson)
\sum_{i=1}^{k} \lambda_i(A+B) \geq \sum_{i=1}^{k} \lambda_i(A) + \sum_{i=1}^{k} \lambda_i(B), in particular substituting the pair A-B,B and noting \lambda_i(A-B) \geq 0 we get the result.

 

[winterfors]

This was exactly what I was looking for, many thanks!